1. **Problem:** Solve the equation $\cos(6x) + 3 \sin(3x) = 2$ for $0 \leq x < 2\pi$. 2. **Recall the range of trigonometric functions:** - $\cos(\theta)$ ranges between $-1$ and $1$. - $\sin(\theta)$ ranges between $-1$ and $1$. 3. **Analyze the equation:** - The maximum value of $\cos(6x)$ is 1. - The maximum value of $3 \sin(3x)$ is 3. - So the maximum possible value of the left side is $1 + 3 = 4$. 4. **Check if the equation can equal 2:** - Since 2 is within the possible range, solutions may exist. 5. **Rewrite the equation:** $$\cos(6x) + 3 \sin(3x) = 2$$ 6. **Use substitution:** Let $y = 3x$, then the equation becomes: $$\cos(2y) + 3 \sin(y) = 2$$ 7. **Express $\cos(2y)$ using double angle formula:** $$\cos(2y) = 1 - 2 \sin^2(y)$$ 8. **Substitute back:** $$1 - 2 \sin^2(y) + 3 \sin(y) = 2$$ 9. **Rearrange:** $$-2 \sin^2(y) + 3 \sin(y) + 1 - 2 = 0$$ $$-2 \sin^2(y) + 3 \sin(y) - 1 = 0$$ 10. **Multiply both sides by -1:** $$2 \sin^2(y) - 3 \sin(y) + 1 = 0$$ 11. **Let $s = \sin(y)$, solve quadratic:** $$2 s^2 - 3 s + 1 = 0$$ 12. **Use quadratic formula:** $$s = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}$$ 13. **Solutions for $s$:** - $s = \frac{3 + 1}{4} = 1$ - $s = \frac{3 - 1}{4} = \frac{2}{4} = 0.5$ 14. **Find $y$ for each $s$:** - For $s = 1$, $\sin(y) = 1 \Rightarrow y = \frac{\pi}{2} + 2k\pi$ - For $s = 0.5$, $\sin(y) = 0.5 \Rightarrow y = \frac{\pi}{6} + 2k\pi$ or $y = \frac{5\pi}{6} + 2k\pi$ 15. **Recall $y = 3x$, solve for $x$:** - For $y = \frac{\pi}{2} + 2k\pi$: $$3x = \frac{\pi}{2} + 2k\pi \Rightarrow x = \frac{\pi}{6} + \frac{2k\pi}{3}$$ - For $y = \frac{\pi}{6} + 2k\pi$: $$3x = \frac{\pi}{6} + 2k\pi \Rightarrow x = \frac{\pi}{18} + \frac{2k\pi}{3}$$ - For $y = \frac{5\pi}{6} + 2k\pi$: $$3x = \frac{5\pi}{6} + 2k\pi \Rightarrow x = \frac{5\pi}{18} + \frac{2k\pi}{3}$$ 16. **Find all $x$ in $[0, 2\pi)$:** - For $x = \frac{\pi}{6} + \frac{2k\pi}{3}$, $k=0,1,2$ gives: - $\frac{\pi}{6}$, $\frac{\pi}{6} + \frac{2\pi}{3} = \frac{5\pi}{6}$, $\frac{\pi}{6} + \frac{4\pi}{3} = \frac{9\pi}{6} = \frac{3\pi}{2}$ - For $x = \frac{\pi}{18} + \frac{2k\pi}{3}$, $k=0,1,2$ gives: - $\frac{\pi}{18}$, $\frac{\pi}{18} + \frac{2\pi}{3} = \frac{13\pi}{18}$, $\frac{\pi}{18} + \frac{4\pi}{3} = \frac{25\pi}{18}$ - For $x = \frac{5\pi}{18} + \frac{2k\pi}{3}$, $k=0,1,2$ gives: - $\frac{5\pi}{18}$, $\frac{5\pi}{18} + \frac{2\pi}{3} = \frac{17\pi}{18}$, $\frac{5\pi}{18} + \frac{4\pi}{3} = \frac{29\pi}{18}$ 17. **Final exact solutions:** $$x = \left\{ \frac{\pi}{18}, \frac{5\pi}{18}, \frac{13\pi}{18}, \frac{17\pi}{18}, \frac{25\pi}{18}, \frac{29\pi}{18}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2} \right\}$$ 18. **Approximate to nearest hundredth:** - $\frac{\pi}{18} \approx 0.17$ - $\frac{5\pi}{18} \approx 0.87$ - $\frac{13\pi}{18} \approx 2.27$ - $\frac{17\pi}{18} \approx 2.97$ - $\frac{25\pi}{18} \approx 4.36$ - $\frac{29\pi}{18} \approx 5.06$ - $\frac{\pi}{6} \approx 0.52$ - $\frac{5\pi}{6} \approx 2.62$ - $\frac{3\pi}{2} \approx 4.71$ **Answer:** The solutions to $\cos(6x) + 3 \sin(3x) = 2$ in $[0, 2\pi)$ are approximately: $$0.17, 0.52, 0.87, 2.27, 2.62, 2.97, 4.36, 4.71, 5.06$$